B

PROPOSITION 8. ALTERNATIVE PROOF.

Let ABC and DEF be two triangles, which have the sides BA, AC equal respectively to the sides ED, DF, and the base BC equal to the base EF.

Then shall the angle BAC be equal to the angle EDF.

For apply the triangle ABC to the triangle DEF, so that B may fall on E, and BC along EF, and so that the point A may be on the side of EF remote from D;

then C must fall on F, since BC is equal to EF. Let GEF be the new position of the triangle ABC. Join DG.

Hence the whole angle EDF = the whole angle EGF; Ax. 2. that is, the angle EDF = the angle BAČ.

Two cases remain which may be dealt with in a similar manner : namely,

QUESTIONS AND EXERCISES FOR REVISION.

1. Define adjacent angles, a right angle, vertically opposite angles. 2. Explain the words enunciation, hypothesis, conclusion.

3. Distinguish between the meanings of the following statements: (i) then AB is equal to PQ;

(ii) then AB shall be equal to PQ.

4. When are two theorems said to be converse to one another. Give an example.

5. Shew by an example that the converse of a true theorem is not itself necessarily true.

6. What is a corollary? Quote the corollary to Proposition 5 ; and shew how its truth follows from that proposition.

7. Name the six parts of a triangle. When are triangles said to be equal in all respects?

8. What do you understand by the expression geometrical magnitudes? Give examples?

9. What is meant by superposition? Explain the test by which Euclid determines if two geometrical magnitudes are equal to one another. Illustrate by an example.

10. Quote and explain the third postulate. What restrictions does Euclid impose on the use of compasses, and what problems are thereby made necessary?

11. Define an axiom. Quote the axioms referred to (i) in Proposition 2; (ii) in Proposition 7.

12. Prove by the method of superposition that two squares are equal in area, if a side of one is equal to a side of the other.

13. Two quadrilaterals ABCD, EFGH have the sides AB, BC, CD, DA equal respectively to the sides EF, FG, GH, HE, and have also the angle BAD equal to the angle FEH. Shew that the figures may be made to coincide with one another.

14. AB, AC are the equal sides of an isosceles triangle ABC; and L, M, N are the middle points of AB, BC, and CA respectively : prove that

(i) LM = MN.

(ii) BN=CL.

(iii) the angle ALM = the angle ANM.

PROPOSITION 9. PROBLEM.

To bisect a given rectilineal angle, that is, to divide it into

two equal parts.

Let BAC be the given angle.
It is required to bisect the angle BAC.

Construction. In AB take any point D;
and from AC cut off AE equal to AD.
Join DE;

I. 3.

remote from A, describe an equi

and on DE, on the side

lateral triangle DEF.

Join AF.

Then shall the straight line AF bisect the angle BAC.

Proof. For in the two triangles DAF, EAF,

Because

DA is equal to EA,

and AF is common to both;

I. 1.

Constr.

and the third side DF is equal to the third side

EF;

Def. 24.

therefore the angle DAF is equal to the angle EAF. I. 8. Therefore the given angle BAC is bisected by the straight line AF.

EXERCISES.

Q.E. F.

1. If in the above figure the equilateral triangle DFE were described on the same side of DE as A, what different cases would arise? And under what circumstances would the construction fail?

2.

In the same figure, shew that AF also bisects the angle DFE. 3. Divide an angle into four equal parts.

PROPOSITION 10. PROBLEM.

To bisect a given finite straight line, that is, to divide it into two equal parts.

Constr.

Let AB be the given straight line.

It is required to divide AB into two equal parts.

On AB describe an equilateral triangle ABC; I. 1. and bisect the angle ACB by the straight line CD, meeting

AB at D.

Then shall AB be bisected at the point D.

Proof. For in the triangles ACD, BCD,

Because

AC is equal to BC,

and CD is common to both;

tained angle BCD;

I. 9.

Def. 24.

also the contained angle ACD is equal to the con

therefore the triangle ACD is equal to the triangle all respects:

Constr. BCD in

I. 4.

so that the base AD is equal to the base BD. Therefore the straight line AB is bisected at the point D.

Q.E.F.

EXERCISES.

1. Shew that the straight line which bisects the vertical angle of an isosceles triangle, also bisects the base.

2.

On a given base describe an isosceles triangle such that the sum of its equal sides may be equal to a given straight line.

PROPOSITION 11. PROBLEM.

To draw a straight line at right angles to a given straight line, from a given point in the same.

Let AB be the given straight line, and C the given point in it.

It is required to draw from C a straight line at right angles

to AB.

Construction. In AC take any point D,

and from CB cut off CE equal to CD.
On DE describe the equilateral triangle DFE.
Join CF.

Proof.

Because

Then shall CF be at right angles to AB.

For in the triangles DCF, ECF,

DC is equal to EC,

and CF is common to both;

I. 3.

I. 1.

Constr.

and the third side DF is equal to the third side

EF:

Def. 24. therefore the angle DCF is equal to the angle ECF: 1. 8. and these are adjacent angles.

But when one straight line, standing on another, makes the adjacent angles equal, each of these angles is called a right angle; Def. 10. therefore each of the angles DCF, ECF is a right angle. Therefore CF is at right angles to AB,

and has been drawn from a point C in it.

EXERCISE.

Q.E.F.

In the figure of the above proposition, shew that any point in FC, or FC produced, is equidistant from D and E.